Question

Решите системы неравенств:
81(а,б)
82(а,б,в)

Answer 1

$left { {{6x^2-29x+30 leq 0} atop {5x+2 textgreater 3x^2}} right.  left { {{6x^2-29x+30 leq 0} atop {5x+2-3x^2 textgreater 0}} right. ; \ 6x^2-29x+30 leq 0; \ D=b^2-4ac; 29^2-4*6*30=121; \ x_1= frac{-b+- sqrt{D} }{2a}; x_1= frac{29-11}{12}=1,5; \ x_2= frac{29+11}{12}= 3 frac{1}{3}$)
x∈[1,5, $3 frac{1}{3}$]
$5x+2-3x^2 textless 0; 3x^2-5x-2 textgreater 0; x_3+x_4= frac{5}{3}; x_3*x_4= -frac{2}{3}; \ x_3=- frac{1}{3}; x_4=2$
x∈(-∞, -1/3)⋃(2, +∞)
Ответ: x∈[1,5;2)

$left { {{x textless 0} atop {21x^2+39x-6 textless 0}} right.; \ 21x^2+39x-6 textless 0 ; 21x^2+39x-6=0; \ x_1+x_2= -frac{b}{a} =- frac{39}{21}; x_1*x_2= frac{c}{a} =- frac{6}{21} ; \ x_1=-2; x_2= frac{1}{7};$
x∈(-2;1/7)
x<0
x∈(-∞, 0)
Ответ: x∈(-2;0)

$left { {{x textgreater 0} atop {4x^2+5x-6 textgreater 0}} right. ; \ 4x^2+5x-6 textgreater 0; 4x^2+5x-6=0; \ x_1+x_2=- frac{b}{a}= -frac{5}{4}; x_1*x_2= frac{c}{a}=- frac{6}{4}; \ x_1=-2; x_2= frac{3}{4};$
x∈(-∞, -2)⋃($frac{3}{4}$, +∞)
x>0
x∈(0, +∞)
Ответ: x∈($frac{3}{4}$;+∞)